Two cars, a and b, are traveling with the same speed of 40. 0 m/s, each having started from rest. Car a has a mass of 1200 kg, a
1 answer:
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Given:
Mass of the rail road car, m = 2 kg
velocity of the three cars coupled system, v' = 1.20 m/s
velocity of first car, = 3 m/s
Solution:
a) Momentum of a body of mass 'm' and velocity 'v' is given by:
p = mv
Now for the coupled system according to law of conservation of momentum, total momentum of a system before and after collision remain conserved:
(1)
where,
= velocity of the first car
= velocity of the 2 coupled cars after collision
Now, from eqn (1)
v' = 1.80 m/s
Therefore, the velocity of the combined car system after collision is 1.80 m/s
Answer:
a) total moment of inertia is 1359.05 kg m^2
b) angular acceleratio is 0.854rad/sec^2
Explanation:
Given data:
m1=6.9 kg
L=4.88 m
m2=34.5 kg
R=1.22 m
we klnow that moment of inertia for rod is given as
J1=(1/12) ×m×L^2
moment of inertia for sphere is given as
J1=(2/5) ×m×r^2
As object rotates around free end of rod then for sphere the axis around what it rotates is at a distance of d2=L+R
For rod distance is d1=0.5*L
By Steiner theorem
for the rod we get
for the sphere we get
And the total moment of inertia for the first case is
b) F=476 N
The torque for system is given as
where a is angle between Force and distance d
and where d represent distance from rotating axis.
In this case a = 90 degree
M=476*2.44 = 1161.44 Nm
The acceleration is calculated as
= 0.854 rad/sec^2